document.write( "Question 913981: Please help me solve this problem:\r
\n" ); document.write( "\n" ); document.write( "The radiator in a car is filled with a solution of 70 per cent antifreeze and 30 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 3.6 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
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Algebra.Com's Answer #554823 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+x+\" = liters of coolant to be drained
\n" ); document.write( "\"+.7x+\" = liters of antifreeze in \"+x+\" liters
\n" ); document.write( "\"+.7%2A3.6+=+2.52+\" = liters of antifreeze
\n" ); document.write( "originally in radiator
\n" ); document.write( "---------------
\n" ); document.write( "Note that \"+x+\" liters drained off are
\n" ); document.write( "replaced with \"+x+\" liters of water, so
\n" ); document.write( "radiator ends up with \"+3.6+\" liters
\n" ); document.write( "---------------
\n" ); document.write( "\"+%28+2.52+-+.7x+%29+%2F+3.6+=+.5+\"
\n" ); document.write( "\"+2.52+-+.7x+=+.5%2A3.6+\"
\n" ); document.write( "\"+2.52+-+.7x+=+1.8+\"
\n" ); document.write( "\"+.7x+=+2.52+-+1.8+\"
\n" ); document.write( "\"+.7x+=+.72+\"
\n" ); document.write( "\"+x+=+1.029+\"
\n" ); document.write( "1.029 liters of coolant must be drained
\n" ); document.write( "off and replaces with water
\n" ); document.write( "--------------------------
\n" ); document.write( "check:
\n" ); document.write( "\"+.7%2A1.029+=+.72+\" liters of coolant drained off
\n" ); document.write( "\"+.3%2A1.029+=+.3087+\" liters of water drained off
\n" ); document.write( "--------------------
\n" ); document.write( "\"+.3%2A3.6+=+1.08+liters+of+water+to+start+with%0D%0A%7B%7B%7B+1.08+-+.3087+=+.7713+\" liters of water remaining
\n" ); document.write( "\"+.7713+%2B+1.029+=+1.8+\" liters of water after
\n" ); document.write( "re-filling with pure water
\n" ); document.write( "\"+3.6+-+1.8+=+1.8+\" liters of antifreeze left
\n" ); document.write( "\"+1.8%2F3.6+=+.5+\"
\n" ); document.write( "OK
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